select i.animal_id, i.namefrom animal_ins i, animal_outs owhere o.animal_id = i.animal_idand i.datetime>o.datetimeorder by i.datetime

select name, datetimefrom (select animal_id, name, datetime from animal_ins where animal_id not in (select animal_id from animal_outs) order by datetime)where rownum

select animal_id, namefrom (select o.animal_id, o.name, (o.datetime-i.datetime) from ANIMAL_OUTS o,ANIMAL_INS i where o.animal_id=i.animal_id order by (o.datetime-i.datetime) desc)where rownum

import java.util.*;class Solution { public ArrayList solution(int[] num_list, int n) { ArrayList answer = new ArrayList(); for(int i=n-1;i

import java.util.*;class Solution { public ArrayList solution(String[] str_list) { ArrayList answer = new ArrayList(); int l = Arrays.asList(str_list).indexOf("l"); int r = Arrays.asList(str_list).indexOf("r"); if(l==-1 && r==-1 ) return answer; if(l==0 && r==str_list.length) return answer; if(r==-1 ||(l=0) ){ for(in..

import java.util.*;class Solution { public ArrayList solution(int[] num_list, int n) { ArrayList answer = new ArrayList(); ArrayList start = new ArrayList(); for(int i=0;i  두개의 ArrayList 합치기addAll() 메소드 사용

SELECT id, name, host_idfrom places awhere (select count(host_id) from places b where a.host_id = b.host_id) >=2order by id

import java.util.*;class Solution { public ArrayList solution(int[] arr, int[][] intervals) { ArrayList answer = new ArrayList(); for(int i=intervals[0][0]; i